New answers tagged bond
3
Greg is quite correct: no clear question. But sometimes you don't know enough to define a question clearly. The implicit question is: "Is this mechanism helpful or even correct?" The second, clear, question, is why does rusting take time to happen.
I think the answer to the first, implicit, question is that it is a start, but far from helpful and ...
4
Getting to know the superoxide anion radical
Hayyan et al. (2016) have studied and characterized the formation of the superoxide anion radical, $\ce{O2^.-}$. This chemical entity forms as a result of the reduction of diatomic oxygen
$\ce{O2 + e^- -> O2^.-}$
There is some conflicting data on the reactivity of this radical, however it can be isolated:
A ...
1
In principle, when studying a structure like HCl, CH3COOH, or any other compound, the first thing to do is drawing the symbol of the atoms in a reasonable geometry on a sheet of paper. Then consider the outer electron layers : put enough points around each symbol to get a reasonable Lewis structure. For example, one point near $\ce{H}$ and six points around $...
-1
Bond energies are important, but you have to start somewhere. Actually, I'm surprised that the bond energies for the four dissociations are so similar. But that's a good start.
Now let's get into a bit more interesting compound: ethane. Very interesting!
The table is only part of the information on ethane given in an open access article (Ref 1). Note that ...
1
On the surface, the answer is no, because that's not what we would call a hydrogen bond.
On a deeper level, the answer is... probably still no for NaH or CsH, because they are too ionic for this. Other hydrides, however, actually do have a thing which looks mighty like an "inverted" hydrogen bond (though we don't normally call it that, either).
...
1
As suggested by @porpyrin, in the second calculation you're assuming the force constant $k$ is constant in both molecules so a ratio can be taken. Let $\tilde{\nu_H}$ and $\tilde{\nu_D}$ be the hydrogen and deuterated versions respectively,
$$\tilde{\nu} = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}}$$
$$\frac{\tilde{\nu_H}}{\tilde{\nu_D}} = \frac{\sqrt{k/\mu_H}}{\...
0
Pauling has calculated the electronegativity of Na by combining the dissociation energy of the three molecules $\ce{H2, NaH, Na2}$. $\ce{Na2}$ is an unstable covalent molecule that do exist in the diluted flames of sodium vapors in contact with chlorine gas at low pressure.
0
VSEPR theory, like VB theory, can be useful but they are quite wrong from first principles. Electrons are indistinguishable, therefore they are delocalized in the entire molecule, once the bonds are formed. In particular, VSEPR theory works for elements of the first and second period, but after that it should be used with care. To fully understand the ...
0
TeF4 is polymeric and low-melting. Not ionic. Covalent.
In general, the elements that are metalloids (on the "border" of metal and non-metal) like Te and Sb, tend to behave chemically like nonmetals (looking at their compounds). But physically as metals (shiny, malleable, conductive). So that's another good clue that these elements would be ...
0
This happens because higher number of fluorine atoms induce a higher positive oxidation state on Xe which causes the Xe atom to shrink in size resulting in a smaller bond length.
XeF2 has Xe in +2 oxidation state while XeF6 has it in +6 oxidation state
link
2
In the harmonic oscillator approximation, the wavenumber of vibration for a diatomic molecule is
$$\tilde{\nu} = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}}$$
Here, $k$ is the force constant of your bond, and $\mu$ is the reduced mass for the two atoms.
$$\mu=\frac{m_1m_2}{m_1+m_2}$$
When you change isotopes, only $\mu$ changes. The force constant depends on the ...
1
In the famous words of George Box, "All models are wrong. Some are useful."
The bent bond and sigma/pi descriptions are two apparently distinct models for representing the underlying structure of electron density of molecules. Both are oversimplifications, but the key question (as with any model) is which is better for accounting for all of the ...
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